//
// Created by Jisam on 2024/8/20 8:34.
// Solution of   #DX0002. 星星
#include <algorithm>
#include <array>
#include <bitset>
#include <cassert>
#include <chrono>
#include <cmath>
#include <cstdint>
#include <cstring>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <vector>
#include <climits>

using namespace std;

#define endl "\n"
#define PSI pair<string,int>
#define PII pair<int,int>
#define PDI pair<double,int>
#define PDD pair<double,double>
#define VVI vector<vector<int>>
#define VI vector<int>
#define VS vector<string>

#define PQLI priority_queue<int, vector<int>, less<int>>
#define PQGI priority_queue<int, vector<int>, greater<int>>
#define code_by_jisam ios::sync_with_stdio(false),cin.tie(nullptr)
using namespace std;
using u32 = unsigned;
using i64 = long long;
using u64 = unsigned long long;
using i128 = __int128;
int dx[] = {-1, 1, 0, 0, 1, 1, -1, -1,};
int dy[] = {0, 0, -1, 1, 1, -1, -1, 1,};

i64 n, k;
const int INF = 1e9;



void solution() {
    cin >> n >> k;
    vector<array<i64,5>> cost(n,{0,0,0,0,0});
    vector<i64> dp(k+1,INF); dp[0]=0; //选i个物品的最低cost
    for (int i = 0; i < n; i++) {
        cin >> cost[i][1] >> cost[i][2] >> cost[i][3]>> cost[i][4] ;
    }
    for(int i = 0; i < n; i ++){
        for(int j = k ;j >= 1; j --)
        {
            for(int l = 0; l <= 4 ; l ++){
                if(j - l >= 0) dp[j] = min(dp[j],dp[j - l] + cost[i][l]);
            }
        }
    }
    cout << dp[k] << endl;
}

int main() {
    code_by_jisam;
    int T = 1;
    cin >> T;
    while (T--) {
        solution();
    }
    return 0;
}